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2 years ago

Solve the inequality7y + 2 < 5y + 12

Mathematics

1 answer:

prisoha [69]2 years ago

7 0

Answer:

y < 5

Step-by-step explanation:

7y + 2 < 5y + 12

-2 -2

7y < 5y + 10

-5y -5y

2y < 10

/2 /2

y < 5

Hope this helps :)

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Suppose xy > 0. Describe the points whose coordinates are solutions to the inequality.

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The solutions fo the inequality are all the points (x, y) that meet these 3 conditions.

x ≠ 0
y ≠ 0
Sign(x) =sign(y)

<h3>Which points are solutions of the inequality?</h3>

We want to find points of the form (x, y) that are solutions of the inequality:

x*y > 0

Clearly x and y must be different than zero.

So, for example if x = -1, y can be any negative number, for example y= -3

x*y > 0

(-1)*(-3) > 0

3 > 0

This is true.

Now if x = 1, y must be positive. LEt's take y = 103, then:

x*y > 0

1*103 > 0

103 > 0

Then we have 3 conditions:

x ≠ 0
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Sign(x) =sign(y)

The solutions fo the inequality are all the points (x, y) that meet these 3 conditions.

If you want to learn more about inequalities:

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lys-0071 [83]

Answer:

Part a) The lateral area is $4r^{2} \pi \ in^{2}$

Part b) The area of the two bases together is $2r^{2} \pi\ in^{2}$

Part c) The surface area is $6r^{2} \pi\ in^{2}$

Step-by-step explanation:

we know that

The surface area of a right cylinder is equal to

$SA=LA+2B$

where

LA is the lateral area

B is the area of the base of cylinder

we have

$r=r\ in$

$h=2r\ in$

Part a) Find the lateral area

The lateral area is equal to

$LA=2\pi rh$

substitute the values

$LA=2\pi r(2r)$

$LA=4r^{2} \pi\ in^{2}$

Part b) Find the area of the two bases together

The area of the base B is equal to

$B=r^{2} \pi\ in^{2}$

the area of the two bases together is

$2B=2r^{2} \pi\ in^{2}$

Part c) Find the surface area of the cylinder

$SA=LA+2B$

we have

$LA=4r^{2} \pi\ in^{2}$

$2B=2r^{2} \pi\ in^{2}$

substitute

$SA=4r^{2} \pi+2r^{2} \pi=6r^{2} \pi\ in^{2}$

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Solve the inequality7y + 2 < 5y + 12​

Solve the inequality7y + 2 < 5y + 12